# Set Theory Question. Trouble defining a function precisely.

1. Oct 7, 2012

### jmjlt88

Let A={1,...,n}. Show that there is a bijection of P(A) with the cartesian product Xn, where X is the two element set X={0,1} and P(A) is the power set of A.

Below is the start of my proof. I just want to make sure that my function "makes sense."

Proof: Let A={1,...n}, and X={0,1}. Define f: P(A) -> Xn by f(A0)=(x1,....,xn), where A0 is a subset of A (and therefore an element of P(A)) and (x1,....,xn) is the element of Xn such that xi=1 if i ε A0 and xi=0 is i is not an element of A0.........

In the next step, I let A0=A1, and show that their image in Xn is the same. I suppose my question is really "how do I ensure f is well-defined?"

2. Oct 8, 2012

### clamtrox

I'd just take an arbitrary element of P(A) and show that it maps to a single element in Xn, then take an arbitrary element in Xn and show exactly one element of P(A) is mapped there. That's enough right?

3. Oct 8, 2012

### HallsofIvy

Staff Emeritus
Given any subset, A, of A, in P(A), assign to every member, x, of A the value "1" if x is in S, "0" if not. That assigns a member of Xn to every member of P(A).